Continuity of the function $\frac{1}{\tan{x}+\sqrt{3}}$ I need to found the continuity of :

$$\frac{1}{\tan{x}+\sqrt{3}}$$

I know the denominator $\ne 0$ but I don't know how to continue and how to clear the variable for find the domain. Thanks!.
 A: Hint:
The domain is defined, as you wrote it, by the condition:
$$\tan x\ne -\sqrt 3=\tan\Bigl(-\frac\pi 3\Bigr)\iff x\ne-\frac\pi 3+k\pi\quad(k\in\mathbf Z).$$
A: As a quotient of two continuous functions, it is continuous wherever it is defined. To find out where that is, we simply need to check where the denominator vanishes, that is, where $\tan x+\sqrt{3}=0$. This is equivalent to $\tan x=-\sqrt{3}$. Can you solve this equation on your own?
A: That's exactly right. The function is going to be continuous everywhere except when the denominator is equal to zero and the points where the function $f(x)=\tan{x}$ is not defined (you simply will not be able to plug them into the tangent function). So, set the denominator equal to zero and solve the following equation for $x$:
$$\tan{x}+\sqrt{3}=0\implies \tan{x}=-\sqrt{3}\implies x=\left \{\frac{2\pi}{3}+2\pi k,\frac{5\pi}{3}+2\pi k, k\in \mathbb{Z}\right \}.$$
The tangent function, $f(x)=\tan{x}$, is not defined on the following set of points: $$x=\left \{\frac{\pi}{2}+\pi k,\ k\in \mathbb{Z}\right \}.$$
So, your original function is defined everywhere expect for the union of those two sets of points.
